Question: Divide the following complex numbers. $ \dfrac{12-18i}{3-3i}$
Solution: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${3+3i}$ $ \dfrac{12-18i}{3-3i} = \dfrac{12-18i}{3-3i} \cdot \dfrac{{3+3i}}{{3+3i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(12-18i) \cdot (3+3i)} {(3-3i) \cdot (3+3i)} = \dfrac{(12-18i) \cdot (3+3i)} {3^2 - (-3i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(12-18i) \cdot (3+3i)} {(3)^2 - (-3i)^2} = $ $ \dfrac{(12-18i) \cdot (3+3i)} {9 + 9} = $ $ \dfrac{(12-18i) \cdot (3+3i)} {18} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({12-18i}) \cdot ({3+3i})} {18} = $ $ \dfrac{{12} \cdot {3} + {-18} \cdot {3 i} + {12} \cdot {3 i} + {-18} \cdot {3 i^2}} {18} $ Evaluate each product of two numbers. $ \dfrac{36 - 54i + 36i - 54 i^2} {18} $ Finally, simplify the fraction. $ \dfrac{36 - 54i + 36i + 54} {18} = \dfrac{90 - 18i} {18} = 5-i $